In the introductory tutorial in
this series we considered a harmonic forcing function **F**_{0}**cos(ωt)**
applied to a translational spring and mass
system with one degree of freedom having mass **m,** spring constant** k**
and a dashpot providing viscous damping with damping constant**
c** shown in Figure 1 below with the spring
extended by distance x .

From the dynamic* free body diagram we obtained the following equation of motion in terms of displacement x and time t

* showing forces producing acceleration of mass m, not static equilibrium

In the previous tutorial we found
the particular solution* to the differential equation of motion
for forced vibrations with a harmonic forcing function F_{0}cos(ωt) without
damping to be:

where ω_{n} is the natural
frequency of the system and x_{0} and v_{0} are the
initial conditions for displacement and velocity respectively.

* the maths tutorial in this series provides an outline of solutions to homogeneous and non-homogeneous second-order ordinary differential equations

The cos(ω_{n}t) and sin(ω_{n}t) terms in equation
(2) express
the free vibration and the cos(ωt) term expresses the steady state motion
attributable to the harmonic forcing function. Because of
damping the
transient cos(ω_{n}t) and sin(ω_{n}t) terms become zero leaving
the cos(ωt) term expressing the steady state vibration.

In this tutorial we are interested in the steady state vibration as this has much greater practical significance. Hence we only consider the particular solution for the non-homogeneous equation of motion (1).

In a previous tutorial we showed
that the general solution to the equation of motion for free
vibrations with damping is a function of cos(ω_{d}t - φ)
where ω_{d} is the damped natural frequency and φ the
phase angle. Thus a reasonable "guess" for the
particular solution to equation (1) for steady state conditions is:

**x**_{p}** =
X.cos(ωt - φ)** ----- (3)

where **X** is the amplitude of the steady state
response and ω the angular frequency of the harmonic forcing
function. There are two unknowns in this equation, X and
φ to be resolved.

Now substitute x_{p}, x_{p}^{/} and x_{p}^{//}
into equation (1)

x_{p} = X.cos(ωt - φ)
x_{p}^{/} = -X.ω.sin(ωt - φ)
x_{p}^{//} = -X.ω^{2}.cos(ωt - φ)
gives:

dividing by k gives:

* see a previous tutorial

We use the following trig identities to create two equations from (3).

*(cos(ωt - φ) = cos(ωt).cos(φ) + sin(ωt).sin(φ)*
and *sin(ωt - φ) = sin(ωt).cos(φ) -
cos(ωt).sin(φ)*

*X.cos(ωt).[A.cos(φ) + B.sin(φ)] + X.sin(ωt).[A.sin(φ) -
B.cos(φ)] = (F*_{0}*/k).cos(ωt)*

which by equivalence of coefficients for cos(ωt) and sin(ωt) terms gives two equations:

*X.cos(ωt).[A.cos(φ) + B.sin(φ)] = (F*_{0}*/k).cos(ωt)
and X.sin(ωt).[A.sin(φ) - B.cos(φ)] = 0*

which reduce to the following equations with unknowns X and φ

*X.[A.cos(φ) + B.sin(φ)] = (F*_{0}*/k)
---- (i) X.[A.sin(φ) - B.cos(φ)] = 0*
---- (ii)

**Summary solution**:

*multiply (i) by cos(φ) and (ii) by sin(φ)* then add the resulting
equations to get *X.A = (F*_{0}*/k).cos(φ)*
---- (iii)

*multiply (i) by sin(φ) and (ii) by cos(φ) *then subtract
resulting (i) - (ii) to get *X.B = (F*_{0}*/k).sin(φ)*
---- (iv)

square both sides of (iii) and (iv) and add the equations to give
* X*^{2}* = (F*_{0}*/k) ^{2}
/ (A^{2} + B^{2})*

substituting back for A and B gives:

Thus the steady state motion of a damped forced vibration with
forcing function F_{0}cos(ωt) is given by equation

**x(t) = X.cos(ωt - φ)** where X
and φ are the above constants.

We now examine the characteristics of the motion in terms of **amplitude
X**, **frequency ratio (ω/ω**_{n})
and **phase angle φ**.

In a previous tutorial we defined the **static amplitude δ = (F _{0
}/ k)** expressing the static
displacement of mass m under force F

Figures 2 and 3 below show the response of the system for two particular sets of parameter values.

Figure 2 shows displacement x(t) for a system where the frequency
ω of the forcing function F_{0}cos(ωt) is less than ω_{n}
the natural frequency of the spring and mass system.
In this example the damping ratio ζ = 0.1 and ω/ω_{n} = 2/3.
The amplification factor is greater than 1. x(t) lags the forcing
function by phase angle φ.

Figure 3 shows displacement x(t) for a system where the frequency
ω of the forcing function F_{0}cos(ωt) is greater than the
natural frequency ω_{n} of the system. In this
example the damping ratio ζ = 0.1 and ω/ω_{n} = 4/3.
The amplification factor is greater than 1. x(t) leads the
forcing function by phase angle φ.

Figures 4 and 5 below illustrate a broader picture of the output characteristics.

In Figure 4 the amplification factor is plotted against the frequency ratio for three values of damping ratio ζ. The amplification ratio peaks at frequency ratio = 1 which is also the resonant condition for the undamped system derived in the previous tutorial. The lower the value of ζ the higher the amplification factor at all frequency ratios. Note that at frequency ratios < 1 the amplification factor is always > 1. As frequency ratios increase beyond the peak value the amplification factor reduces progressively to become < 1 and ultimately tends to zero. These characteristics are similar to the undamped system.

Note the marked points for frequency ratios used in Figures 2 and 3 for ζ = 0.1.

Figure 5 shows the variation of phase angle (units in radians) with frequency ratio for three values of damping ratio ζ. The principal characteristics are:

- At frequency ratios < 1 displacement x(t) lags the forcing function and the phase angle increases as the frequency ratio increases.
- When the frequency ratio = 1 the phase angle is 90°.
- At frequency ratios >1 displacement x(t) leads the forcing function and the phase angle increases as the frequency ratio increases, tending to 180° . This characteristic is more pronounced the lower the damping ratio.

Note the marked points for frequency ratios used in Figures 2 and 3 for ζ = 0.1.

The phase angle "flips" when ratio ω/ω_{n}
transitions through the value 1. Figures 6 and 7 below show
phase angles respectively at frequency ratios just less than 1 and
just greater than 1.

This tutorial concludes the series on mechanical vibrations for the present. I plan to continue the series to cover torsional vibrations, multiple degree of freedom systems and transverse vibration of beams.

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